Trigonometry Class 10 Solved Problems and Examples, Prove Identities by Given Angle

Master Class 10 Trigonometry with easy-to-understand solved problems and examples. Perfect for CBSE exams and competitive test (JEE, NDA) preparation by Anand Classes. Boost your exam scores with Anand Classes’ expert solutions.

If A = 30^o, then Prove That:

(i) sin 2A = 2sin A cos A = 2 tan A/(1 + tan²A)

(ii) cos 2A = cos²A – sin²A = (1 – tan²A)/(1 + tan²A)

(iii) 2 cos² A – 1 = 1 – 2 sin²A

(iv) sin 3A = 3 sin A – 4 sin³A.

Solution:

(i) Given A = 30^o

Sin 2A = sin 2(30^o)

= sin 60^o = √3/2

2 sin A cos A = 2 sin 30^o cos 30^o

= 2 (½) (√3/2) = √3/2

Now,

2 tan A/(1 + tan²A) = 2 tan 30⁰/(1 + tan² 30⁰)

= 2(1/√3)/(1 +(1/√3)²) = √3/2

(ii) cos 2A = cos 2 (30⁰) = cos 60^o = ½

Cos² A – sin² A = cos² 30^o – sin² 30^o

= 3/4 – 1/4 = 1/2

(1 – tan²A)/(1 + tan²A) = (1 – tan²30⁰)/(1 + tan²30⁰)

= (1 -(1/√3)²)/(1 +(1/√3)²) = 1/2

(iii) 2 cos² A – 1 = 2 cos² 30^o – 1

= 2 (¾) – 1 = 3/2 – 1 = ½

1 – 2 sin² A = 1 – 2 sin² 30^o

= 1 – 2 (¼) = ½

2 cos² A – 1 = 1 – sin² A

(iv) sin 3A = sin 3 (30^o)

= sin 90^o = 1

3 sin A – 4 sin³ A = 3 sin 30^o – 4 sin³ 30^o

= 3 (½) – 4 (½)³ = 3/2 – ½ = 1

Therefore,

Sin 3A = 3 sin A – 4 sin³ A

If A = B = 45^o, show that:

(i) sin (A – B) = sin A cos B – cos A sin B

(ii) cos (A + B) = cos A cos B – sin A sin B

Solution:

Given that A = B = 45^o

(i) LHS = sin (A – B)

= sin (45^o – 45^o)

= sin 0^o

= 0

RHS = sin A cos B – cos A sin B

= sin 45^o cos 45^o – cos 45^o sin 45^o

= 1/√2 (1/√2) – 1/√2 (1/√2)

= 0

Therefore, LHS = RHS

(ii) LHS = cos (A + B)

= cos (45^o + 45^o)

= cos 90^o

= 0

RHS = cos A cos B – sin A sin B

= cos 45^o cos 45^o – sin 45^o sin 45^o

= 1/√2 (1/√2) – 1/√2 (1/√2)

= 0

Therefore, LHS = RHS

If A = 30^o; show that:

(i) sin 3 A = 4 sin A sin (60^o – A) sin (60^o + A)

(ii) (sin A – cos A)² = 1 – sin 2A

(iii) cos 2A = cos⁴ A – sin⁴ A

(iv) (1 – cos 2A)/sin 2A = tan A

(v) (1 + sin 2 A + cos 2 A) / (sin A + cos A) = 2 cos A.

(vi) 4 cos A cos (60^o – A). cos (60^o + A) = cos 3A

(vii) (cos³A – cos 3A)/cos A + (sin³A + sin 3A)/sin A = 3

Solution:

Given that A = 30^o

(i) According to the question we have

LHS = sin 3A

= sin 3 (30^o)

= sin 90^o

= 1

RHS = 4 sin A sin (60^o – A) sin (60^o + A)

= 4 sin A sin (60^o – 30^o) sin (60^o + 30^o)

= 4 (½) (½) (1)

= 1

LHS = RHS

(ii) According to the question we have

LHS = (sin A – cos A)²

= (sin 30^o – cos 30^o)²

= (½ – √3/2)²

= ¼ + ¾ – √3/2

= 1 – √3/2

= (2 – √3)/2

RHS = 1 – sin 2A

= 1 – sin 2 (30^o)

= 1 – sin 60^o

= 1 – √3/2

= (2 – √3)/2

Therefore, LHS = RHS

(iii) According to the question we have

LHS = cos 2A

= cos 2 (30^o)

= cos 60^o

= ½

RHS = cos⁴ A – sin⁴ A

= cos⁴ 30^o – sin⁴ 30^o

= (√3/2)⁴ – (½)⁴

= 9/16 – 1/16

= ½

LHS = RHS

(iv) According to the question we have

LHS = (1 – cos 2A)/ sin 2A

= 1 – cos 2(30^o)/ sin 2 (30^o)

= 1 – ½/ (√3/2)

= 1/√3

RHS = tan A

= tan 30^o

= 1/√3

LHS = RHS

(v) According to the question we have

$$
\text{LHS} = \frac{1 + \sin 2A + \cos 2A}{\sin A + \cos A}
$$
$$
= \frac{1 + \sin 2(30^\circ) + \cos 2(30^\circ)}{\sin 30^\circ + \cos 30^\circ}
$$
$$
= \frac{1 + \frac{\sqrt{3}}{2} + \frac{1}{2}}{\frac{1}{2} + \frac{\sqrt{3}}{2}}
$$
$$
= \frac{1 + \frac{\sqrt{3}}{2} + \frac{1}{2}}{\frac{1}{2} + \frac{\sqrt{3}}{2}}
$$
$$
= \frac{3 + \sqrt{3}}{\sqrt{3} + 1} \times \frac{\sqrt{3} – 1}{\sqrt{3} – 1}
$$
$$
= \frac{3\sqrt{3} – 3 + 3 – \sqrt{3}}{2}
$$
$$
= \frac{2\sqrt{3}}{2}
$$
$$
= \sqrt{3}
$$

RHS = 2 cos A

= 2 cos 30^o

= 2 (√3/2)

= √3

(vi) According to the question we have

LHS = 4 cos A cos (60^o – A) cos (60^o + A)

= 4 cos A cos (60^o – 30^o) cos (60^o + 30^o)

= 4 cos 30^o cos 30^o cos 90^o

= 4 (√3/2) (√3/2) 0

= 0

RHS = cos 3A

= cos 3 (30^o)

= cos 90^o

= 0

LHS = RHS

(vii) According to the question we have

$$
\text{LHS} = \frac{\cos^3 A – \cos 3A}{\cos A} + \frac{\sin^3 A + \sin 3A}{\sin A}
$$
$$
= \frac{\cos^3 30^\circ – \cos 3(30^\circ)}{\cos 30^\circ} + \frac{\sin^3 30^\circ + \sin 3(30^\circ)}{\sin 30^\circ}
$$
$$
= \frac{\left( \frac{\sqrt{3}}{2} \right)^3 – 0}{\frac{\sqrt{3}}{2}} + \frac{\left( \frac{1}{2} \right)^3 + 1}{\frac{1}{2}}
$$
$$
= \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}} + \frac{9}{8} \times \frac{1}{\frac{1}{2}}
$$
$$
= \left( \frac{\sqrt{3}}{2} \right)^2 + \frac{9}{4}
$$

= ¾ + 9/4

= 12/4

= 3 = RHS

Hence the proof

Do You Know?

• The values of trigonometric ratios are exact for 0°, 30°, 45°, 60°, and 90°.
• These ratios are frequently used in geometry, physics, astronomy, and engineering.
• The identities like sin²θ + cos²θ = 1 help derive all other formulas in trigonometry.

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